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        2021年4月24日 凌晨
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            <h1 style="display: none">动态规划</h1>
            
            <div class="markdown-body">
              <p>本文主要是写题时遇到的动态规划问题的整理</p>
<h1 id="动态规划"><a href="#动态规划" class="headerlink" title="动态规划"></a>动态规划</h1><h2 id="1-背包问题"><a href="#1-背包问题" class="headerlink" title="1.背包问题"></a>1.背包问题</h2><p>背包问题是一种组合优化的 NP 完全问题：</p>
<p>​    <strong>有 N 个物品和容量为 W 的背包，每个物品都有 自己的体积w和价值v，求拿哪些物品可以使得背包所装下物品的总价值最大。</strong></p>
<p>如果限定每种物品只能选择0个或1个，则问题称为0-1背包问题；如果不限定每种物品的数量，则问题称为无界背包问题或完全背包问题。</p>
<h3 id="0-1背包"><a href="#0-1背包" class="headerlink" title="0-1背包"></a>0-1背包</h3><p>​    以<strong>0-1背包问题</strong>为例。我们可以定义一个二维数组dp 存储最大价值，其中dp[i] [j]示前i件物品体积不超过j的情况下能达到的最大价值。在我们遍 历到第i件物品时，在当前背包总容量为j的情况下，如果我们不将物品i放入背包，那么<br>$$<br>dp[i][j] =dp[i-1][j]<br>$$<br>，即前 i 个物品的最大价值等于只取前 i-1 个物品时的最大价值；如果我们将物品 i 放 入背包，假设第i件物品体积为w，价值为v，那么我们得到dp[i] [j]=dp[i-1] [j-w]+v。我们只需 在遍历过程中对这两种情况取最大值即可，总时间复杂度和空间复杂度都为O(NW)。</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br></pre></td><td class="code"><pre><code class="hljs c++"><span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">knapsack</span><span class="hljs-params">(vector&lt;<span class="hljs-keyword">int</span>&gt; weights, vector&lt;<span class="hljs-keyword">int</span>&gt; values, <span class="hljs-keyword">int</span> N, <span class="hljs-keyword">int</span> W)</span> </span><br><span class="hljs-function"></span>&#123; vector&lt;vector&lt;<span class="hljs-keyword">int</span>&gt;&gt; <span class="hljs-built_in">dp</span>(N + <span class="hljs-number">1</span>, vector&lt;<span class="hljs-keyword">int</span>&gt;(W + <span class="hljs-number">1</span>, <span class="hljs-number">0</span>)); <br>  <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> i = <span class="hljs-number">1</span>; i &lt;= N; ++i) <br>  &#123; <br>      <span class="hljs-keyword">int</span> w = weights[i<span class="hljs-number">-1</span>], v = values[i<span class="hljs-number">-1</span>]; <br>      <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> j = <span class="hljs-number">1</span>; j &lt;= W; ++j) <br>      &#123; <br>          <span class="hljs-keyword">if</span> (j &gt;= w) <br>          &#123; <br>              dp[i][j] = <span class="hljs-built_in">max</span>(dp[i<span class="hljs-number">-1</span>][j], dp[i<span class="hljs-number">-1</span>][j-w] + v); <br>          &#125; <br>          <span class="hljs-keyword">else</span> <br>          &#123; <br>              dp[i][j] = dp[i<span class="hljs-number">-1</span>][j]; <br>          &#125; <br>      &#125; <br>  &#125; <br> <span class="hljs-keyword">return</span> dp[N][W]; <br>&#125;<br><br></code></pre></td></tr></table></figure>

<p><img src="E:\nodejs\blog\source\img\image-20210424100051871.png" srcset="/img/loading.gif" lazyload></p>
<p><strong>空间优化</strong></p>
<p>在程序实现时可以对 0-1 背包做优化。观察状态转移方程可以知道，前 i 件物品的状态仅与前 i-1 件物品的状态有关，因此可以将 dp 定义为一维数组，其中 dp[j] 既可以表示 dp[i-1] [j] 也可以表示 dp[i] [j]。此时，</p>
<p>​<br>$$<br>dp[j]=max(ma(dp[j],dp[j-w]+v))<br>$$<br>因为 dp[j-w] 表示 dp[i-1] [j-w]，因此不能先求 dp[i] [j-w]，防止将 dp[i-1] [j-w] 覆盖。也就是说要先计算 dp[i] [j] 再计算 dp[i] [j-w]，在程序实现时需要按<strong>倒序来循环</strong>求解。</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><code class="hljs c++"><span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">knapsack</span><span class="hljs-params">(vector&lt;<span class="hljs-keyword">int</span>&gt; weights, vector&lt;<span class="hljs-keyword">int</span>&gt; values, <span class="hljs-keyword">int</span> N, <span class="hljs-keyword">int</span> W)</span> </span><br><span class="hljs-function"></span>&#123; <span class="hljs-function">vector&lt;<span class="hljs-keyword">int</span>&gt; <span class="hljs-title">dp</span><span class="hljs-params">(W + <span class="hljs-number">1</span>, <span class="hljs-number">0</span>)</span></span>; <br>  <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> i = <span class="hljs-number">1</span>; i &lt;= N; ++i) <br>  &#123; <br>      <span class="hljs-keyword">int</span> w = weights[i<span class="hljs-number">-1</span>], v = values[i<span class="hljs-number">-1</span>]; <br>      <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> j = W; j &gt;= w; --j) <br>      &#123; <br>          dp[j] = <span class="hljs-built_in">max</span>(dp[j], dp[j-w] + v); <br>      &#125; <br>  &#125; <br>  <span class="hljs-keyword">return</span> dp[W];<br>&#125;<br></code></pre></td></tr></table></figure>



<h3 id="完全背包"><a href="#完全背包" class="headerlink" title="完全背包"></a>完全背包</h3><p>在完全背包问题中，一个物品可以拿多次。</p>
<p>假设我们遍历到物品 i = 2， 且其体积为w=2，价值为v=3；对于背包容量j=5，最多只能装下2个该物品。那么我们的状 态转移方程就变成了<br>$$<br>dp[2][5]=max(dp[1][5],dp[1][3]+3,dp[1][1]+6)<br>$$<br>如果采用这种方法，假设背包容量无穷大而物体的体积无穷小，我们这里的比较次数也会趋近于无穷大，远超 O(NW)的时间复杂度。 怎么解决这个问题呢？我们发现在 dp[2] [3] 的时候我们其实已经考虑了 dp[1] [3] 和 dp[2] [1] 的情况，而在时 dp[2] [1]也已经考虑了 dp[1] [1] 的情况。因此，如图下半部分所示，对于拿多个物品的情况，我们只需考虑 dp[2] [3] 即可，即<br>$$<br>dp[2] [5] = max(dp[1] [5], dp[2] [3] + 3)<br>$$<br>这样，我们 就得到了完全背包问题的状态转移方程：dp[i] [j]=max(dp[i-1] [j],dp[i] [j-w]+v)，其与0-1背包问题的差别仅仅是把状态转移方程中的第二个i-1变成了i。</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br></pre></td><td class="code"><pre><code class="hljs c++"><span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">knapsack</span><span class="hljs-params">(vector&lt;<span class="hljs-keyword">int</span>&gt; weights, vector&lt;<span class="hljs-keyword">int</span>&gt; values, <span class="hljs-keyword">int</span> N, <span class="hljs-keyword">int</span> W)</span> </span><br><span class="hljs-function"></span>&#123; <br>    vector&lt;vector&lt;<span class="hljs-keyword">int</span>&gt;&gt; <span class="hljs-built_in">dp</span>(N + <span class="hljs-number">1</span>, vector&lt;<span class="hljs-keyword">int</span>&gt;(W + <span class="hljs-number">1</span>, <span class="hljs-number">0</span>)); <br>    <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> i = <span class="hljs-number">1</span>; i &lt;= N; ++i) <br>    &#123; <br>        <span class="hljs-keyword">int</span> w = weights[i<span class="hljs-number">-1</span>], v = values[i<span class="hljs-number">-1</span>]; <br>        <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> j = <span class="hljs-number">1</span>; j &lt;= W; ++j) <br>        &#123; <br>            <span class="hljs-keyword">if</span> (j &gt;= w) <br>            &#123; <br>                dp[i][j] = <span class="hljs-built_in">max</span>(dp[i<span class="hljs-number">-1</span>][j], dp[i][j-w] + v); <br>            &#125; <br>            <span class="hljs-keyword">else</span> <br>            &#123; <br>                dp[i][j] = dp[i<span class="hljs-number">-1</span>][j]; <br>            &#125; <br>        &#125; <br>    &#125; <br>    <span class="hljs-keyword">return</span> dp[N][W]; <br>&#125; <br></code></pre></td></tr></table></figure>

<p><img src="E:\nodejs\blog\source\img\image-20210424101145002.png" srcset="/img/loading.gif" lazyload></p>
<p><strong>空间优化</strong></p>
<p>同样的，我们也可以利用空间压缩将时间复杂度降低为 O(W)。这里要注意我们在遍历每一行的时候必须<strong>正向遍历</strong>，因为我们需要利用当前物品在第j-w列的信息。</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br></pre></td><td class="code"><pre><code class="hljs c++"><span class="hljs-function"><span class="hljs-keyword">int</span> <span class="hljs-title">knapsack</span><span class="hljs-params">(vector&lt;<span class="hljs-keyword">int</span>&gt; weights, vector&lt;<span class="hljs-keyword">int</span>&gt; values, <span class="hljs-keyword">int</span> N, <span class="hljs-keyword">int</span> W)</span> </span><br><span class="hljs-function"></span>&#123; <span class="hljs-function">vector&lt;<span class="hljs-keyword">int</span>&gt; <span class="hljs-title">dp</span><span class="hljs-params">(W + <span class="hljs-number">1</span>, <span class="hljs-number">0</span>)</span></span>; <br>  <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> i = <span class="hljs-number">1</span>; i &lt;= N; ++i) <br>  &#123; <br>      <span class="hljs-keyword">int</span> w = weights[i<span class="hljs-number">-1</span>], v = values[i<span class="hljs-number">-1</span>]; <br>      <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> j = w; j &lt;= W; ++j) <br>      &#123; <br>          dp[j] = <span class="hljs-built_in">max</span>(dp[j], dp[j-w] + v); <br>      &#125; <br>  &#125; <br>  <span class="hljs-keyword">return</span> dp[W]; <br>&#125; <br></code></pre></td></tr></table></figure>






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